2022 Edexcel AS Level Biology B (8BI0) 模擬試題連答案詳解

To calculate the mitotic index:
1. Identify the total number of cells undergoing mitosis (prophase, metaphase, anaphase, and telophase):
\(45 + 12 + 8 + 15 = 80\) cells.
2. Divide the number of mitotic cells by the total number of observed cells:
\(\frac{80}{400} = 0.20\).
3. Multiply by 100 to convert to a percentage:
\(0.20 \times 100 = 20.0\\%\).

1 mark for the correct answer: B (20.0%).

Incorrect options:
- A (80.0%) is the percentage of cells in interphase (\(\frac{320}{400} \times 100\)).
- C (11.25%) is only the percentage of cells in prophase (\(\frac{45}{400} \times 100\)).
- D (3.75%) is only the percentage of cells in telophase (\(\frac{15}{400} \times 100\)).

Active loading of sucrose involves the active transport of hydrogen ions (\(\text{H}^+\)) out of the companion cells into the cell wall space using ATP. This establishes a high concentration of protons in the cell wall. Protons then diffuse back into the companion cells down their concentration gradient through co-transporter proteins, carrying sucrose molecules with them against their concentration gradient. Sucrose then diffuses into the sieve tube elements via plasmodesmata.

1 mark for the correct answer: B.

Incorrect explanations:
- A is incorrect because sucrose is not directly transported out of companion cells using ATP.
- C is incorrect because sucrose enters companion cells against its concentration gradient via active co-transport, not simple diffusion.
- D is incorrect because hydrogen ions are pumped out of the companion cells, not the sieve tube elements, to drive the co-transport of sucrose.

When the left ventricle begins to contract, the ventricular pressure rises. As soon as the pressure in the ventricle exceeds the pressure in the atrium, the atrioventricular (bicuspid) valve closes to prevent backflow of blood. The semilunar valve remains closed because the ventricular pressure is still lower than the aortic pressure. This short phase of contraction with all valves closed is known as isovolumetric contraction.

1 mark for the correct answer: B.

Incorrect options:
- A is incorrect because the semilunar valve only opens once ventricular pressure exceeds the pressure in the aorta.
- C is incorrect because the atrioventricular valve closes rather than opens.
- D is incorrect because the semilunar valve is already closed during diastole and remains closed during this phase.

The described features (peptidoglycan cell wall, 70S ribosomes, circular naked DNA, and plasmids) identify the cell as a prokaryotic bacterium. Prokaryotic cells lack any membrane-bound organelles such as mitochondria, rough endoplasmic reticulum, or Golgi apparatus. However, some prokaryotes possess flagella for locomotion.

1 mark for the correct answer: C.

Incorrect options:
- A, B, and D are membrane-bound organelles which are exclusively found in eukaryotic cells and are absent in prokaryotic organisms.

Cellulose is an unbranched polymer of \(\beta\)-glucose molecules joined by \(\beta\)-1,4-glycosidic bonds. To allow these bonds to form, successive \(\beta\)-glucose molecules must be rotated by \(180^\circ\) relative to each other. This specific orientation allows hydrogen bonds to form between parallel cellulose chains, resulting in the formation of strong microfibrils.

1 mark for the correct answer: D.

Incorrect options:
- A describes the structure of amylose, which is made of \(\alpha\)-glucose molecules and forms a helix.
- B describes a highly branched structure (similar to amylopectin or glycogen) but incorrectly states it is made of \(\beta\)-glucose.
- C is incorrect because cellulose is made of \(\beta\)-glucose, not \(\alpha\)-glucose.

At high temperatures (above \(60^\circ\text{C}\)), proteins embedded in the cell membranes (including the plasma membrane and tonoplast) denature. In addition, the kinetic energy of the phospholipids increases, significantly increasing membrane fluidity and disrupting the integrity of the bilayer. This makes the membranes highly permeable, allowing the water-soluble pigment (betalain) to escape rapidly into the surrounding solution by diffusion.

1 mark for the correct answer: B.

Incorrect options:
- A is incorrect because high temperatures denature transport proteins, which halts any active transport processes.
- C is incorrect because membranes do not remain intact or stable at these high temperatures.
- D is incorrect because the rapid release of pigment is caused by structural damage to the membranes, not by the chemical breakdown (hydrolysis) of the pigment molecule.

The Bohr effect causes the oxygen dissociation curve of hemoglobin to shift to the right when carbon dioxide levels rise (resulting in a lower blood pH). A shift to the right indicates that at any given partial pressure of oxygen (\(p\text{O}_2\)), hemoglobin has a lower affinity for oxygen and will release (unload) its oxygen more easily to supply active, respiring tissues that are producing carbon dioxide.

1 mark for the correct answer: B.

Incorrect options:
- A is incorrect because shifting the curve to the left would increase oxygen affinity, making it harder for oxygen to be unloaded in respiring tissues.
- C is incorrect because the Bohr effect facilitates, rather than prevents, the unloading of oxygen under physiological conditions.
- D is incorrect because the Bohr effect is concerned with the oxygen affinity changes in response to carbon dioxide and pH, and has no relation to carbon monoxide.

First, calculate the total number of cells undergoing mitosis: \(18 + 12 + 6 + 4 = 40\) cells. Next, divide the number of mitotic cells by the total number of cells observed: \(40 \div 240 \approx 0.1667\). Converting to two significant figures gives \(0.17\) (or \(17\%\)).

1 mark: Correct calculation of total mitotic cells (40).
1 mark: Division of mitotic cells by total cells (40/240).
0.5 marks: Correct answer rounded to 2 significant figures (0.17 or 17%).

Water molecules are polar and form hydrogen bonds with each other. This causes cohesion, allowing water molecules to be pulled up in a continuous, unbroken column. Water molecules also form hydrogen bonds with the hydrophilic components (cellulose and lignin) of the xylem walls, known as adhesion, which helps prevent the water column from dropping due to gravity. Tension generated by transpiration pulls this cohesive column of water upwards.

1 mark: Reference to cohesion as hydrogen bonding between water molecules forming a continuous column.
1 mark: Reference to adhesion as hydrogen bonding between water molecules and xylem walls.
0.5 marks: Explaining that transpiration pulls this column of water upwards under tension.

Arteries have a much thicker layer of smooth muscle and elastic fibers in their walls (tunica media) compared to veins. This structural design is essential because arteries carry blood directly from the heart under high, fluctuating pressure; the elastic fibers allow the artery to stretch and recoil to maintain blood pressure, while the smooth muscle can contract to control blood flow. In contrast, veins carry blood under low pressure back to the heart, so they have thinner walls with less muscle and elastic fibers, and instead have valves to prevent backflow.

1 mark: Stating that arteries have a thicker layer of elastic fibers and smooth muscle than veins.
1 mark: Explaining that elastic recoil and muscle contraction in arteries are needed to withstand and maintain high pressure.
0.5 marks: Stating that veins have thinner walls because they carry blood under much lower pressure.

Ribosomes: Eukaryotic cells contain larger 80S ribosomes (in the cytoplasm and rough endoplasmic reticulum), whereas prokaryotes contain smaller 70S ribosomes. DNA organization: Eukaryotic DNA is linear, wrapped around histone proteins, and enclosed within a membrane-bound nucleus. Prokaryotic DNA is circular, 'naked' (not associated with histones), and located freely in the cytoplasm in a region called the nucleoid, often accompanied by small plasmids.

1 mark: Ribosome comparison (80S in eukaryotes vs 70S in prokaryotes).
1 mark: DNA shape/association comparison (linear with histones in eukaryotes vs circular and naked/no histones in prokaryotes).
0.5 marks: Location comparison (enclosed in a nucleus in eukaryotes vs free in the cytoplasm/nucleoid in prokaryotes).

First, a spontaneous mutation occurs in the bacterial genome, giving rise to an allele that confers resistance to a specific antibiotic. When the antibiotic is applied, it acts as a strong selection pressure. Non-resistant bacteria are killed, whereas the resistant mutant bacteria survive. These surviving bacteria reproduce rapidly by binary fission, passing the resistance allele to their offspring, thereby increasing the frequency of the resistance gene in the population over time.

1 mark: Random mutation occurs producing a resistant allele.
1 mark: Antibiotic acts as a selection pressure, killing non-resistant bacteria and allowing resistant bacteria to survive.
0.5 marks: Surviving bacteria reproduce (by binary fission) and pass on the resistant allele, increasing its frequency in the population.

Similarity: Both amylose and glycogen are polymers made of \(\alpha\)-glucose monomers linked by \(\alpha\)-1,4-glycosidic bonds. Differences: (1) Glycogen is highly branched containing \(\alpha\)-1,6-glycosidic bonds, while amylose is unbranched and contains only \(\alpha\)-1,4-glycosidic bonds. (2) Amylose coils into a tight helical structure, whereas glycogen's frequent branching prevents it from forming a tight helix, making it highly compact and easily hydrolysed.

1 mark: Identifying the similarity (both are polymers of alpha-glucose with 1,4-glycosidic bonds).
1 mark: Identifying Difference 1 (glycogen is branched with 1,6-glycosidic bonds, while amylose is unbranched).
0.5 marks: Identifying Difference 2 (amylose forms a helical shape, whereas glycogen does not due to branching).

Similarity: Both active transport and facilitated diffusion rely on specific transmembrane proteins (carrier proteins) to transport molecules that cannot easily cross the phospholipid bilayer directly. Differences: (1) Active transport moves molecules against their concentration gradient (from a low concentration to a high concentration), whereas facilitated diffusion moves molecules down their concentration gradient (from a high concentration to a low concentration). (2) Active transport requires metabolic energy in the form of ATP, whereas facilitated diffusion is a completely passive process that relies solely on the kinetic energy of the particles.

1 mark: Identifying the similarity (both require transmembrane/carrier proteins to transport molecules).
1 mark: Identifying Difference 1 (active transport goes against the concentration gradient, whereas facilitated diffusion goes down the concentration gradient).
0.5 marks: Identifying Difference 2 (active transport requires ATP, whereas facilitated diffusion is passive).

During vigorous exercise, muscle cells respire at a high rate, producing large amounts of carbon dioxide (\(CO_2\)). This \(CO_2\) dissolves in the blood to form carbonic acid, which dissociates into hydrogen ions (\(H^+\)) and hydrogencarbonate ions, lowering the pH. This increased acidity causes a conformational change in hemoglobin, reducing its affinity for oxygen and shifting the oxygen dissociation curve to the right (the Bohr effect). This is highly significant as it allows oxygen to be more easily unloaded from hemoglobin at the respiring muscle tissues where it is urgently needed.

1 mark: Explaining that high respiration rates release more CO2, decreasing the pH of the blood.
1 mark: Stating that lower pH reduces hemoglobin's affinity for oxygen, shifting the dissociation curve to the right.
0.5 marks: Explaining that this allows oxygen to be more readily released/unloaded to actively respiring tissues.

The three-domain system was proposed based on molecular phylogenetics, specifically the comparison of ribosomal RNA (rRNA) nucleotide sequences. Woese showed that life should be classified into three distinct domains: Archaea, Bacteria, and Eukaryota. He found that Archaea, despite being single-celled prokaryotes, are biochemically and genetically distinct from Bacteria (e.g., they lack peptidoglycan and have different membrane lipid bonds) and share key genetic transcription mechanisms with Eukaryotes.

1 mark for identifying that classification is based on molecular phylogeny / ribosomal RNA sequencing. 1 mark for specifying that the three domains are Archaea, Bacteria, and Eukaryota, showing Archaea are distinct from Bacteria. 0.5 marks for noting biochemical differences such as the presence of peptidoglycan in bacterial cell walls but not in Archaea.

1. Calculate the total number of dividing cells: \(24 + 12 + 6 + 6 = 48\) cells. 2. Calculate the mitotic index by dividing dividing cells by the total number of cells: \(\frac{48}{160} = 0.3\). 3. Convert to a percentage: \(0.3 \times 100 = 30\%\). 4. A high mitotic index indicates a high rate of cell division, which is a hallmark of cancerous or malignant tissues.

1 mark for showing correct addition and division: \(\frac{48}{160}\). 1 mark for the correct percentage of 30% (or 30.0%). 0.5 marks for stating that a high mitotic index indicates rapid, uncontrolled cell division characteristic of malignant cancers.

Water evaporates from the cell walls of mesophyll cells in leaves (transpiration), lowering water potential and creating a tension that pulls the water column upwards. Cohesion, driven by hydrogen bonding between individual water molecules, forms a continuous, unbroken column of water through the xylem vessels. Adhesion of water molecules to the hydrophilic cellulose walls of the xylem vessels prevents the water column from pulling away or collapsing under gravity.

1 mark for explaining that transpiration from leaves creates tension that pulls water up. 1 mark for describing cohesion due to hydrogen bonding forming a continuous water column. 0.5 marks for noting adhesion of water molecules to the xylem cell walls.

Triglycerides consist of three fatty acid chains ester-linked to a single glycerol molecule, making them entirely hydrophobic and non-polar, suitable for energy storage. Phospholipids contain only two fatty acid chains and a highly polar, hydrophilic phosphate group attached to the glycerol. This amphipathic nature enables phospholipids to spontaneously form a stable bilayer in water, which forms the structural basis of cell membranes.

1 mark for contrasting composition: triglycerides have three fatty acids and no phosphate, whereas phospholipids have two fatty acids and a polar phosphate group. 1 mark for explaining that phospholipids are amphipathic with hydrophilic heads and hydrophobic tails, forming a bilayer. 0.5 marks for linking this bilayer to the selective barrier function of cell membranes.

The acrosome reaction begins when receptors on the sperm head bind to glycoproteins on the zona pellucida of the oocyte. This causes the sperm cell membrane and the outer acrosome membrane to fuse, releasing hydrolytic enzymes (such as acrosin) via exocytosis. These enzymes digest a path through the jelly-like zona pellucida, enabling the sperm membrane to reach and fuse with the oocyte plasma membrane, allowing the sperm nucleus to enter.

1 mark for explaining that binding to the zona pellucida triggers fusion of the sperm plasma and outer acrosomal membranes. 1 mark for describing the release of hydrolytic enzymes by exocytosis to digest the zona pellucida. 0.5 marks for stating the purpose is to allow the sperm plasma membrane to reach and fuse with the oocyte plasma membrane.

An increase in carbon dioxide concentration reacts with water to form carbonic acid, which lowers blood pH. This increased acidity alters the tertiary structure of hemoglobin, reducing its affinity for oxygen and shifting the oxygen dissociation curve to the right (the Bohr effect). This ensures that in actively respiring tissues, where carbon dioxide concentration is high, oxygen is unloaded more readily from hemoglobin to meet the high demand of aerobic respiration.

1 mark for stating that increased carbon dioxide concentration shifts the oxygen dissociation curve to the right (Bohr shift). 1 mark for explaining that increased carbon dioxide lowers pH, which alters hemoglobin's structure and reduces its affinity for oxygen. 0.5 marks for explaining that this promotes the release of oxygen to actively respiring tissues.

Amylose is an unbranched, helical chain of \(\alpha\)-glucose monomers linked only by \(\alpha\)-1,4-glycosidic bonds, making it highly compact. Amylopectin is a highly branched polymer containing \(\alpha\)-1,4-glycosidic bonds along its main chain and \(\alpha\)-1,6-glycosidic bonds at branch points. This highly branched structure creates many terminal glucose ends that can be simultaneously targeted by amylase enzymes, allowing rapid hydrolysis into glucose when energy is urgently required.

1 mark for distinguishing structures: amylose is unbranched with alpha-1,4-bonds, while amylopectin is branched with both alpha-1,4 and alpha-1,6-glycosidic bonds. 1 mark for explaining that the branched structure provides multiple terminal ends for simultaneous enzymatic hydrolysis. 0.5 marks for linking this to rapid glucose mobilization for respiration.

Plant cell walls are composed of cellulose, which consists of long, unbranched chains of \(\beta\)-glucose linked by \(\beta\)-1,4-glycosidic bonds that form hydrogen-bonded microfibrils embedded in pectin and hemicellulose. Gram-positive bacterial cell walls consist of a thick, rigid layer of peptidoglycan (murein) composed of repeating disaccharides cross-linked by short peptide chains, along with teichoic acids. Plant cell walls do not contain peptidoglycan and are unaffected by antibiotics like penicillin, which target peptidoglycan cross-linking.

1 mark for contrasting chemical units: plant cell walls contain cellulose (beta-glucose microfibrils) whereas Gram-positive bacterial cell walls contain peptidoglycan (amino sugars cross-linked by peptides). 1 mark for structural differences: Gram-positive walls are thick, single-layered peptidoglycan structures containing teichoic acids. 0.5 marks for noting that bacterial walls are targeted by antibiotics such as penicillin which inhibit transpeptidase enzymes, unlike plant walls.

To calculate the mitotic index, use the formula:
\(\text{Mitotic Index} = \frac{\text{Number of cells in mitosis}}{\text{Total number of cells}} \times 100\)

1. Find the total number of cells undergoing mitosis (prophase + metaphase + anaphase + telophase):
\(28 + 12 + 6 + 4 = 50\) cells.

2. Divide by the total number of cells observed (160) and multiply by 100:
\(\frac{50}{160} \times 100 = 31.25\%\).

- 1 mark: Correct calculation of the total number of dividing cells (50) OR showing the correct formula setup: \(\frac{50}{160}\).
- 1 mark: Correct calculation of the percentage as 31.25% (or rounded to 31.3%).
- 0.5 marks: Stating the correct unit (%) with the final answer.

Lignin is a strong, impermeable polymer deposited in the secondary cell walls of xylem vessels. It provides mechanical strength to withstand the high negative pressure (tension) generated by transpiration, preventing the vessel elements from collapsing inward. Its hydrophobic nature waterproofs the walls, ensuring efficient longitudinal water transport. Depositing lignin in patterns such as spirals or rings (annular) rather than a solid uniform block leaves unlignified areas (pits) for lateral transport and allows the xylem vessels to flex and stretch during plant elongation and growth.

- 1 mark: For stating that lignin provides mechanical strength to prevent the vessel collapsing under tension/negative pressure.
- 1 mark: For explaining that lignin waterproofs the vessel walls, ensuring water is retained inside the lumen for transport.
- 0.5 marks: For explaining that spiral, annular, or ring-like deposition patterns allow flexibility, elongation, or growth of the stem.

The left ventricle has a much thicker muscular wall than the right ventricle, allowing it to contract more forcefully. This generates the high hydrostatic pressure needed to overcome high resistance and pump blood through the systemic circulation to reach all body tissues. In contrast, the right ventricle only pumps blood to the lungs (pulmonary circulation). The lungs are located close to the heart, presenting low resistance; therefore, a much lower pressure is sufficient, which also protects the delicate capillaries in the lungs from bursting.

- 1 mark: Explaining that the left ventricle requires high pressure to overcome the high resistance of the systemic circulation to pump blood around the entire body.
- 1 mark: Explaining that the right ventricle pumps blood through the shorter, lower-resistance pulmonary circulation to the lungs.
- 0.5 marks: Pointing out that lower pressure in the pulmonary circulation prevents damage/rupturing of the delicate pulmonary capillaries or alveoli.

Both amylose and amylopectin are polymers made up of \(\alpha\)-glucose monomers. However, they differ structurally:
1. Amylose consists of an unbranched chain of glucose units linked solely by \(\alpha\)-1,4-glycosidic bonds, which coils into a compact helix.
2. Amylopectin is a highly branched molecule containing \(\alpha\)-1,4-glycosidic bonds along its main chain, with \(\alpha\)-1,6-glycosidic bonds forming branch points approximately every 24 to 30 glucose units.

- 1 mark: For identifying amylose as unbranched/helical containing only \(\alpha\)-1,4-glycosidic bonds.
- 1 mark: For identifying amylopectin as branched containing both \(\alpha\)-1,4 and \(\alpha\)-1,6-glycosidic bonds.
- 0.5 marks: For noting that both are composed of \(\alpha\)-glucose monomers.

1. Ribosomes: Prokaryotes contain smaller 70S ribosomes, while eukaryotic animal cells contain larger 80S ribosomes in the cytoplasm (and rough endoplasmic reticulum).
2. Genetic Material: Prokaryotic DNA is circular, 'naked' (not associated with histone proteins), and floats freely in the cytoplasm within the nucleoid region (along with smaller plasmids). Eukaryotic DNA is linear, associated with histones to form chromatin, and enclosed within a membrane-bound nucleus.

- 1 mark: Distinguishing ribosome size: prokaryotes have 70S ribosomes vs. eukaryotes have 80S ribosomes.
- 1 mark: Distinguishing DNA structure: prokaryotes have circular/naked DNA, whereas eukaryotes have linear DNA associated with histone proteins.
- 0.5 marks: Distinguishing DNA location: prokaryotic DNA is free-floating in the cytoplasm/nucleoid, whereas eukaryotic DNA is contained within a nucleus.

The tertiary structure of a globular protein is stabilized by interactions between R-groups, including hydrogen bonds, ionic bonds, disulfide bridges (covalent bonds), and hydrophobic interactions.
Among these, ionic bonds are highly sensitive to pH. A change in pH alters the concentration of \(\text{H}^+\) or \(\text{OH}^-\)` ions, which changes the charge on the acidic or basic R-groups (e.g., carboxyl or amino groups). This disrupts the electrostatic attractions (ionic bonds) holding the tertiary shape together, leading to denaturation. Hydrogen bonds are also disrupted by changes in pH.

- 1 mark: Correctly naming any two bonds that stabilize tertiary structure (e.g., hydrogen bonds, ionic bonds, disulfide bridges, or hydrophobic interactions).
- 1 mark: Stating that ionic bonds (or hydrogen bonds) are the most sensitive to changes in pH.
- 0.5 marks: Explaining that pH changes alter the charge / protonation of amino acid R-groups, disrupting these electrostatic attractions.

Fick's Law of Diffusion states that:
\(\text{Rate of Diffusion} \propto \frac{\text{Surface Area} \times \text{Difference in Concentration}}{\text{Thickness of Membrane}}

Mammalian alveoli are adapted in the following ways:
1. Surface Area: Millions of alveoli present a massive surface area for gas exchange.
2. Thickness of Membrane: The barrier is extremely thin. Both the alveolar wall (squamous epithelium) and the capillary wall are only one cell thick, minimizing the diffusion distance.
3. Concentration Difference: Continuous ventilation (breathing) and constant blood flow in the adjacent capillaries maintain a steep concentration gradient for oxygen and carbon dioxide.

- 1 mark: Stating Fick's Law accurately in words or as a formula showing that rate is proportional to surface area and concentration gradient, and inversely proportional to membrane thickness.
- 1 mark: Linking alveolar structures to at least two variables in the equation: huge surface area (many alveoli) AND short diffusion distance (one-cell-thick walls).
- 0.5 marks: Linking continuous ventilation or blood flow to maintaining a steep concentration gradient.

Collagen structural features and relation to function: 1. It is a fibrous protein with structural function. 2. Consists of three polypeptide chains wound around each other to form a tight triple helix (tropocollagen). 3. Has a repeating sequence where every third amino acid is glycine. Glycine is the smallest amino acid (hydrogen R-group), allowing the chains to pack very closely together. 4. Covalent cross-links form between adjacent collagen molecules to form fibrils, providing high tensile strength and resistance to stretching. 5. It is insoluble in water, maintaining structural integrity. Comparison with haemoglobin structure: 1. Haemoglobin is a globular protein with metabolic/transport functions, whereas collagen is fibrous and structural. 2. Haemoglobin consists of four polypeptide chains (two alpha and two beta subunits) forming a spherical shape, whereas collagen has three polypeptide chains in a long helix. 3. Haemoglobin is soluble in water because its hydrophilic R-groups face outwards and hydrophobic R-groups face inwards, whereas collagen is insoluble. 4. Haemoglobin contains a non-protein prosthetic group (haem group containing \(Fe^{2+}\)) in each subunit to bind oxygen, whereas collagen does not contain prosthetic groups.

Level 1 (1-2 marks): Simple description of either collagen or haemoglobin structure, or basic comparative statements without linking structure to function. Level 2 (3-4 marks): Detailed description of collagen structure (e.g., triple helix, glycine, cross-links) and how this relates to its high tensile strength OR clear structural comparison with haemoglobin (e.g., fibrous vs globular, insoluble vs soluble). Level 3 (5-6 marks): Comprehensive comparison showing detailed structural adaptations of collagen (glycine allowing tight packing, covalent cross-links forming fibrils, insolubility) directly contrasted with haemoglobin (quaternary structure of four subunits, soluble due to outer hydrophilic groups, presence of haem/\(Fe^{2+}\) prosthetic groups).

Water uptake and pathway across the root: 1. Water enters root hair cells from the soil by osmosis, moving down a water potential gradient. 2. Water moves across the root cortex via two main pathways: the apoplast pathway (through cell walls/intercellular spaces) and the symplast pathway (through cytoplasm and plasmodesmata). 3. When water reaches the endodermis, its progress via the apoplast pathway is blocked by the waterproof Casparian strip (made of suberin). 4. This forces water to cross the selectively permeable cell membrane into the symplast pathway, allowing the plant to control which ions enter the xylem. Movement up the stem (transpiration stream): 5. Transpiration (evaporation of water vapor) occurs from the leaves through stomata, lowering the water potential of mesophyll cells. 6. Water moves out of the xylem into the mesophyll cells, creating a tension (negative pressure) at the top of the xylem vessel. 7. Cohesion between water molecules (due to hydrogen bonding) forms a continuous, unbroken column of water from root to leaf. 8. Adhesion of water molecules to the polar lignin walls of xylem vessels prevents the column from breaking and supports it against gravity, pulling the entire column upwards (cohesion-tension theory).

Level 1 (1-2 marks): Identifies osmosis as the entry mechanism or names the apoplast/symplast pathways or mentions cohesion/adhesion. Level 2 (3-4 marks): Explains how water crosses the root (contrasting apoplast/symplast and describing the block at the Casparian strip) OR explains the cohesion-tension theory of transpiration pull up the xylem. Level 3 (5-6 marks): Provides a complete and detailed explanation of both phases: describes root transport (osmosis, apoplast/symplast pathways, role of endodermis/Casparian strip) and stem transport (evaporation creating tension, cohesion keeping the water column continuous, adhesion to xylem walls supporting the column).

First, calculate the total number of cells observed: 120 + 45 + 30 + 15 + 790 = 1000 cells. The proportion of cells in metaphase is 45 / 1000 = 0.045. Convert the total cell cycle duration from hours to minutes: 18 hours * 60 minutes/hour = 1080 minutes. Finally, calculate the duration of metaphase: 0.045 * 1080 = 48.6 minutes.

1 mark for the correct answer A. Method: (45 / 1000) * 1080 = 48.6 minutes.

Active loading involves hydrogen ions (protons) being actively pumped out of the companion cells into the cell wall space using ATP. This creates a proton gradient. Protons then diffuse back into the companion cells down their concentration gradient through a co-transporter protein, bringing sucrose molecules with them against the sucrose concentration gradient. Sucrose then diffuses into the sieve tube elements via plasmodesmata.

1 mark for the correct answer A.

For blood to be ejected from the left ventricle into the aorta, the pressure in the left ventricle must exceed the pressure in the aorta. This occurs during ventricular systole (specifically during the ejection phase) when the semi-lunar valve is forced open.

1 mark for the correct answer C.

The presence of peptidoglycan in the cell wall, 70S ribosomes, and circular naked DNA are key diagnostic features of the domain Bacteria. Archaea have cell walls lacking peptidoglycan, and Eukaryota have 80S ribosomes and linear DNA associated with histones.

1 mark for the correct answer B.

Directional selection occurs when environmental character changes select for an extreme phenotype (banded shells) over the other (unbanded shells), causing a directional shift in allele frequencies over time.

1 mark for the correct answer C.

Cellulose consists of \(\beta\)-glucose monomers joined by 1,4-glycosidic bonds, which requires alternating glucose molecules to be rotated 180 degrees. Amylose is an unbranched polymer of \(\alpha\)-glucose monomers joined by 1,4-glycosidic bonds that coils into a helix.

1 mark for the correct answer A.

Fick's Law states that Rate of Diffusion is proportional to \(\frac{\text{Surface Area} \times \text{Concentration Difference}}{\text{Thickness of membrane}}\). Therefore, doubling the surface area directly doubles the rate of diffusion.

1 mark for the correct answer C.

The Bohr effect states that an increase in \(p\text{CO}_2\) causes the oxygen haemoglobin dissociation curve to shift to the right. This shift decreases haemoglobin's affinity for oxygen, making it easier for oxygen to dissociate and diffuse into the respiring tissues that need it.

1 mark for the correct answer B.

The wave of electrical excitation originates in the sinoatrial node (SAN), which acts as the pacemaker. It spreads across the muscle walls of both atria, stimulating atrial systole. A band of non-conductive tissue prevents the wave from passing directly into the ventricles. Instead, it reaches the atrioventricular node (AVN), which introduces a short delay (approx. 0.1s). This delay is crucial as it allows the atria to fully contract and empty their blood into the ventricles before ventricular systole begins. The excitation wave is then conducted rapidly down the septum via the Bundle of His and through the Purkyne fibres, causing the ventricles to contract simultaneously from the apex upwards, pushing blood out into the arteries.

Award 0.5 marks for each point up to a maximum of 2.5 marks:
1. Wave of excitation originates at the sinoatrial node (SAN) / pacemaker.
2. Spreads across atrial walls causing atrial contraction / atrial systole.
3. Reaches the atrioventricular node (AVN) which delays the electrical impulse.
4. Delay allows atria to completely empty/ventricles to fill before ventricles contract.
5. Impulse passes down the Bundle of His / Purkyne fibres to initiate ventricular contraction from the apex.

Xylem vessels are highly specialized for water transport. They are composed of dead cells aligned end-to-end. The end walls break down entirely to form continuous, hollow tubes, and the lack of cytoplasm or organelles provides an unobstructed pathway for water flow. The cell walls are heavily thickened with a tough polymer called lignin. Lignin provides exceptional mechanical strength, which prevents the vessel walls from collapsing inward when subjected to high negative pressure (tension) as water is pulled up the plant. Additionally, non-lignified areas called pits allow lateral water movement between adjacent vessels.

Award 0.5 marks for each point up to a maximum of 2.5 marks:
1. Dead cells aligned end-to-end with no end walls to form a continuous/hollow tube.
2. Lack of cytoplasm / cell contents reduces resistance to water flow.
3. Cell walls are thickened / impregnated with lignin.
4. Lignin provides mechanical strength / prevents inward collapse of walls under tension/negative pressure.
5. Pits (unlignified regions) allow lateral water movement between adjacent vessels.

Mitosis must produce genetically identical daughter cells. In metaphase, chromosomes fully condense and align individually along the equator of the spindle. Spindle fibres attach to the centromere of each chromosome from opposite poles. This precise alignment ensures that when the centromere splits during anaphase, the genetically identical sister chromatids are pulled apart by the shortening spindle fibres to opposite poles. Because sister chromatids are exact replicates of each other, this separation ensures that each resulting pole, and subsequently each daughter cell, receives an identical set of chromosomes with no loss or gain of genetic material.

Award 0.5 marks for each point up to a maximum of 2.5 marks:
1. Metaphase: Chromosomes align individually along the equator / metaphase plate of the spindle.
2. Spindle fibres attach to the centromeres from opposite poles.
3. Anaphase: Centromeres divide / split, separating the sister chromatids.
4. Shortening spindle fibres pull sister chromatids to opposite poles of the cell.
5. Since sister chromatids are genetically identical, this ensures both daughter cells receive an exact/identical copy of the genetic material.

Simpson's Index of Diversity is calculated as \(D = 1 - \sum \left(\frac{n}{N}\right)^2\) or \(D = 1 - \frac{\sum n(n-1)}{N(N-1)}\), where \(n\) is the number of individuals of a particular species and \(N\) is the total number of organisms of all species. A high value of \(D\) (close to 1) represents high species diversity, which means there is high species richness and evenness. High-diversity communities are ecologically stable, meaning they can better resist environmental disturbances. A low value of \(D\) (close to 0) indicates low diversity, where the community is dominated by only one or a few species; such habitats are unstable and minor environmental changes can cause drastic changes to the whole ecosystem.

Award 0.5 marks for each point up to a maximum of 2.5 marks:
1. Correct formula: \(D = 1 - \sum \left(\frac{n}{N}\right)^2\) OR \(D = 1 - \frac{\sum n(n-1)}{N(N-1)}\).
2. Correctly defines \(n\) as the number of individuals of a single species and \(N\) as the total number of individuals of all species.
3. High \(D\) value indicates high species diversity / richness and evenness.
4. High \(D\) indicates a stable community that is less susceptible to environmental change / pest outbreaks.
5. Low \(D\) value indicates low diversity / dominated by a few species, making the ecosystem unstable / vulnerable.

Carbon dioxide released by respiring tissues diffuses into red blood cells. Inside the cell, \(CO_2\) reacts with water (\(H_2O\)) to form carbonic acid (\(H_2CO_3\)), a reaction catalysed by the enzyme carbonic anhydrase. Carbonic acid is highly unstable and rapidly dissociates into hydrogen ions (\(H^+\)) and hydrogencarbonate ions (\(HCO_3^-\)). The hydrogencarbonate ions then diffuse out of the red blood cells down their concentration gradient into the blood plasma, where they are transported to the lungs. To maintain electrical neutrality inside the red blood cell, chloride ions (\(Cl^-\)) diffuse into the cell from the plasma, a process known as the chloride shift.

Award 0.5 marks for each point up to a maximum of 2.5 marks:
1. Carbon dioxide diffuses into red blood cells (erythrocytes).
2. \(CO_2\) reacts with water to form carbonic acid (\(H_2CO_3\)), catalysed by carbonic anhydrase.
3. Carbonic acid dissociates to release hydrogen ions (\(H^+\)) and hydrogencarbonate ions (\(HCO_3^-\)).
4. Hydrogencarbonate (\(HCO_3^-\)) ions diffuse out of the red blood cell into the blood plasma.
5. Chloride ions (\(Cl^-\)) diffuse into the red blood cell to maintain electrical neutrality (chloride shift).

Natural selection operates on pre-existing genetic variation in a population, which arises from random gene mutations. When a new selection pressure is introduced (such as a new predator, climate change, or a disease), individuals with certain phenotypes have a selective advantage. These individuals are more likely to survive and successfully reproduce (differential reproductive success) than those without the advantageous phenotype. They pass on their advantageous alleles to their offspring. Over successive generations, the frequency of the advantageous allele increases in the population's gene pool, leading to evolutionary change.

Award 0.5 marks for each point up to a maximum of 2.5 marks:
1. Genetic variation exists within the population due to random gene mutations.
2. Introduction of a new selection pressure makes certain phenotypes/alleles advantageous.
3. Individuals with these advantageous alleles/phenotypes have a higher survival rate (survival of the fittest).
4. Surviving individuals are more likely to reproduce and pass on their advantageous alleles to their offspring.
5. Over generations, the frequency of the advantageous allele increases in the population's gene pool.

Plant cells and bacterial cells have distinct structural differences: 1) Eukaryotic plant cells contain their DNA inside a double-membrane bound nucleus, whereas prokaryotic bacteria have naked, circular DNA free in the cytoplasm (the nucleoid). 2) Plant cells contain 80S ribosomes and membrane-bound organelles like mitochondria and chloroplasts, whereas bacteria have 70S ribosomes and lack all membrane-bound organelles. 3) Plant cell walls are composed of cellulose, whereas bacterial cell walls are composed of peptidoglycan (murein). The function of the cellulose cell wall in plant cells is to provide mechanical strength and structural support, keeping the cell turgid and preventing osmotic lysis when water enters.

Award 0.5 marks for each point up to a maximum of 2.5 marks:
1. Difference 1: Plant cells have a nucleus enclosing linear DNA, whereas bacteria have circular DNA free in the cytoplasm/nucleoid.
2. Difference 2: Plant cells have 80S ribosomes / membrane-bound organelles (e.g. mitochondria, chloroplasts), whereas bacteria have 70S ribosomes / lack membrane-bound organelles.
3. Difference 3: Plant cell walls are made of cellulose, whereas bacterial cell walls are made of peptidoglycan / murein.
4. Difference 4: Bacteria can contain plasmids, whereas plant cells do not.
5. Function: Cell wall provides mechanical support / structural strength / maintains turgidity / prevents osmotic lysis (accept any one valid function).

Both active transport and facilitated diffusion are specialized transport mechanisms across biological membranes. They share similarities: both utilize transmembrane transport proteins (such as carrier proteins) to assist the movement of polar, charged, or large molecules that cannot cross the hydrophobic lipid bilayer on their own. However, they differ significantly in energetics and direction: facilitated diffusion is a passive process that relies on the kinetic energy of particles to move them down their concentration gradient (from high to low concentration) without requiring metabolic energy (ATP). In contrast, active transport moves substances against their concentration gradient (from low to high concentration) and therefore requires energy in the form of ATP, which is hydrolysed to drive the transport proteins.

Award 0.5 marks for each point up to a maximum of 2.5 marks:
1. Similarity: Both require specific membrane/transmembrane proteins (carrier proteins) to transport substances across the bilayer.
2. Similarity: Both transport polar, charged, or large molecules (which cannot cross the phospholipid bilayer directly).
3. Difference: Active transport moves substances against their concentration gradient, whereas facilitated diffusion moves substances down their concentration gradient.
4. Difference: Active transport requires ATP / metabolic energy, whereas facilitated diffusion is a passive process / does not require ATP.
5. Difference: Active transport relies solely on carrier proteins, whereas facilitated diffusion can use both channel and carrier proteins.

1. Find the radius of the capillary tube:
\(r = \frac{0.8\text{ mm}}{2} = 0.4\text{ mm}\)

2. Calculate the volume of water taken up (cylinder volume):
\(V = \pi r^2 h\)
\(V = 3.14 \times (0.4)^2 \times 45\)
\(V = 3.14 \times 0.16 \times 45\)
\(V = 22.608\text{ mm}^3\)

3. Calculate the rate of water uptake per minute:
\(\text{Rate} = \frac{22.608\text{ mm}^3}{15\text{ min}} = 1.5072\text{ mm}^3\,\text{min}^{-1}\)

4. Round to 2 significant figures:
\(1.5\text{ mm}^3\,\text{min}^{-1}\)

- 1 mark: Correct calculation of the volume of the capillary cylinder (\(22.6\text{ mm}^3\) or \(22.61\text{ mm}^3\)) using the correct radius of 0.4 mm.
- 1 mark: Division of the volume by 15 minutes to find the rate (\(1.51\) or \(1.5072\)).
- 0.5 marks: Correctly rounded final answer to 2 significant figures: \(1.5\text{ mm}^3\,\text{min}^{-1}\).

Upon contact with the zona pellucida, the acrosome membrane fuses with the sperm cell surface membrane, releasing digestive enzymes (such as acrosin) via exocytosis. These enzymes digest the zona pellucida pathway, allowing the sperm to reach and fuse with the oocyte's plasma membrane. Contact triggers the cortical reaction, in which cortical granules in the oocyte fuse with the plasma membrane and release enzymes into the zona pellucida. This hardens the glycoprotein matrix, forming a fertilization membrane that prevents any further sperm from entering.

- 1 mark: Correct description of the acrosome reaction, highlighting the release of hydrolytic enzymes to digest the zona pellucida.
- 1 mark: Reference to the cortical reaction and the release of cortical granules by exocytosis.
- 0.5 marks: Explanation of the hardening of the zona pellucida to form an impenetrable barrier (fertilisation membrane) to block polyspermy.

Similarity: Both amylose and cellulose are polysaccharides composed of glucose monomers linked by 1,4-glycosidic bonds.

Differences:
1. Monomer: Amylose consists of \(\alpha\)-glucose monomers, whereas cellulose consists of \(\beta\)-glucose monomers.
2. Rotation: In cellulose, alternate glucose monomers are rotated \(180^\circ\) to allow 1,4-linkages, whereas they are in the same orientation in amylose.
3. Shape & Function: Amylose forms a coiled, compact helical structure making it highly suitable for energy storage in plants. Cellulose forms straight, unbranched chains that run parallel to one another, forming cross-linked hydrogen bonds to create strong microfibrils that provide structural support to plant cell walls.

- 1 mark: Contrast in monomer units (\(\alpha\)-glucose in amylose vs \(\beta\)-glucose in cellulose) and/or the \(180^\circ\) rotation of alternate monomers in cellulose.
- 1 mark: Contrast in structural shape and function (coiled helical structure for compact storage in amylose vs straight unbranched chains for structural support in cellulose).
- 0.5 marks: Mention of hydrogen bonds forming between adjacent cellulose chains to form microfibrils.

As carbon dioxide concentration increases, it dissolves in the blood to form carbonic acid, which dissociates into hydrogen ions (\(H^+\)) and hydrogencarbonate ions, lowering the blood pH. The increased acidity alters the tertiary structure of haemoglobin, reducing its affinity for oxygen. Consequently, the oxygen dissociation curve shifts to the right (the Bohr effect), which facilitates the dissociation/unloading of oxygen at actively respiring tissues where carbon dioxide concentrations are highest.

- 1 mark: States that the curve shifts to the right (Bohr effect).
- 1 mark: Explains that higher \(CO_2\) leads to lower pH/higher hydrogen ion concentration, which changes haemoglobin's tertiary structure and reduces its affinity for oxygen.
- 0.5 marks: Relates this to the functional advantage of releasing/unloading more oxygen to actively respiring tissues.

1. Find the total number of cells: \(192 + 22 + 10 + 6 + 10 = 240\) cells.
2. Calculate the proportion of cells in metaphase: \(\frac{10}{240} = \frac{1}{24}\).
3. Convert the total cycle duration to minutes: \(18\text{ hours} \times 60\text{ minutes/hour} = 1080\text{ minutes}\).
4. Calculate the duration of metaphase: \(\frac{1}{24} \times 1080\text{ minutes} = 45\text{ minutes}\).

- 1 mark: Calculates the correct proportion of cells in metaphase (\(\frac{10}{240}\) or \(0.0417\) / \(4.17\%\)).
- 1 mark: Converts 18 hours to 1080 minutes.
- 0.5 marks: Multiplies proportion by total minutes to give the final correct answer of 45 minutes.

1. Change in environment: Reduced vegetation cover resulted in a lighter background, making dark banded snails easier for predators (e.g., birds) to spot.
2. Survival advantage: Light unbanded snails had a higher level of camouflage, providing a selective advantage (higher survival rate).
3. Reproduction and inheritance: Surviving light unbanded snails reproduced and passed on their advantageous alleles for light/unbanded shells to their offspring.
4. Allele frequency: Over generations, this selection pressure increased the frequency of the light unbanded phenotype and its corresponding alleles within the gene pool.

- 1 mark: Identifies the selection pressure (predation) and explains why light unbanded snails are better camouflaged/adapted in the modified environment.
- 1 mark: Connects survival to reproduction and the inheritance of the advantageous allele(s) by the offspring.
- 0.5 marks: States that this process leads to an increase in the allele frequency of the advantageous allele over generations.

The aorta must withstand and maintain high blood pressure as blood leaves the heart. It is adapted by having a thick layer of elastic fibres (elastin) that stretch to accommodate high-pressure blood during ventricular systole, and recoil to maintain diastolic pressure. It also has thick collagen layers to prevent bursting. Conversely, capillaries are adapted for fast exchange of materials; their walls are composed of a single layer of squamous endothelial cells, providing an extremely short diffusion pathway for nutrients and gases.

- 1 mark: Outlines the adaptation of the aorta (thick elastic wall/elastin to stretch and recoil, or collagen to withstand pressure).
- 1 mark: Outlines the adaptation of a capillary (walls that are only one cell thick/squamous endothelium to minimize diffusion distance).
- 0.5 marks: Explicitly connects these structures to their biological roles (sustaining high systemic pressure vs enabling rapid material exchange).

Observable Differences (any two):
1. Mitochondria possess an inner membrane folded into cristae, whereas chloroplasts have an internal system of flat thylakoid discs stacked into grana.
2. Chloroplasts may contain visible starch granules or plastoglobules, which are absent in mitochondria.
3. Chloroplasts are typically larger and have a different overall shape compared to mitochondria.

Observable/Structural Similarities (any one):
1. Both organelles are surrounded by a double membrane (an outer and inner membrane).
2. Both contain 70S ribosomes (seen as small dots in the matrix/stroma).
3. Both contain circular loops of DNA (stroma/matrix).
4. Both have fluid-filled compartments (matrix in mitochondria, stroma in chloroplasts).

- 1 mark: One valid structural difference observed via TEM (e.g., cristae vs thylakoids/grana).
- 1 mark: A second valid structural difference observed via TEM (e.g., presence of starch grains in chloroplasts).
- 0.5 marks: One valid structural similarity (e.g., double membrane envelope, fluid-filled interior, or presence of 70S ribosomes).

Hydrogen ions are actively pumped out of companion cells into the cell wall using ATP, establishing a proton gradient. These ions diffuse back into the companion cells down their concentration gradient via co-transporter proteins, bringing sucrose with them against its concentration gradient. Sucrose then diffuses into the sieve tube element via plasmodesmata. This accumulation of sucrose lowers the water potential, causing water to enter from the xylem by osmosis. This creates high hydrostatic pressure at the source. At the sink, sucrose is unloaded, water potential increases, water leaves, lowering hydrostatic pressure. Sap flows from high to low hydrostatic pressure.

1. Active transport of hydrogen ions / protons out of companion cells using ATP. (1 mark) 2. Hydrogen ions diffuse back down concentration gradient via co-transporter proteins, bringing sucrose in. (1 mark) 3. Sucrose moves into sieve tube element via plasmodesmata. (1 mark) 4. Presence of sucrose lowers water potential, causing water to move in by osmosis from the xylem. (1 mark) 5. This increases hydrostatic pressure at the source. (1 mark) 6. Unloading of sucrose at sink decreases pressure, establishing a hydrostatic pressure gradient driving mass flow. (1 mark)

At the arteriole end, the hydrostatic pressure of blood is higher than the oncotic pressure of plasma proteins, resulting in a net outward filtration pressure that forces water and solutes out to form tissue fluid. Large proteins remain in the capillary. At the venule end, hydrostatic pressure drops significantly due to resistance, becoming lower than the oncotic pressure. Consequently, water is drawn back into the capillary by osmosis. In a protein-deficient diet, liver synthesis of plasma proteins decreases, lowering plasma protein concentration. This reduces blood oncotic pressure, meaning less tissue fluid is reabsorbed at the venule end, causing tissue fluid to accumulate in the intercellular space (edema).

1. Arteriole end: Hydrostatic pressure is greater than oncotic pressure, forcing water and small solutes out of the capillary. (1 mark) 2. Large proteins and red blood cells cannot pass through and remain in the blood. (1 mark) 3. Venule end: Hydrostatic pressure drops below oncotic pressure, causing net movement of water back into the capillary by osmosis. (1 mark) 4. Protein deficiency reduces plasma protein synthesis in the liver. (1 mark) 5. This lowers the oncotic pressure of blood inside the capillaries. (1 mark) 6. Less water is reabsorbed at the venule end, leading to tissue fluid accumulation / swelling / edema. (1 mark)

To ensure unbiased sampling, the ecologist should use a grid system and random coordinates generated by a computer to place the pitfall traps. All traps must be left for the same duration to standardise the method. Simpson's Index of Diversity \(D\) can be calculated using the formula \(D = 1 - \sum \left(\frac{n}{N}\right)^2\), where n represents the number of individuals of each species and N is the total number of individuals of all species. D values range from 0 to 1, with values closer to 1 representing high biodiversity, which includes both high species richness and high species evenness. A high value of \(D\) indicates a stable ecosystem because complex food webs are present. If one species is affected by a change, such as disease, predators can feed on alternative prey species, meaning the community is less likely to be disrupted.

1. Use of a coordinate grid and random number generator to place pitfall traps randomly. (1 mark) 2. Standardisation of sampling technique (e.g. traps left for the same time / same size trap). (1 mark) 3. Calculation formula components: n is number of individuals of a single species and N is total number of individuals of all species. (1 mark) 4. High D indicates high species richness (number of different species) and species evenness (relative abundance of each). (1 mark) 5. High diversity is associated with complex food webs and more ecological niches. (1 mark) 6. This leads to greater ecosystem stability because there is alternative prey/food sources if one species declines. (1 mark)